While I was learning about Langmuir's adsorption isotherm in my chemistry class, my teacher talked about the situation when the rate of adsorption and desorption will become equal. However, he mentioned that the rate of adsorption is dependent on pressure but the rate of desorption is not and when he gave an explanation I wasn't satisfied with it. I would imagine that greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface. But that isn't what my teacher said. Please explain.


Without knowing exactly what you were told its hard to be exact, but here is a description of the situation.

The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one does not depend on the other, i.e if one site is occupied it does not affect adjacent sites an any way. Note that this is a model and this determines how we look at the problem. The model is then $$M(g) + S(surface) = GM(surface) $$ where G is gas and S surface and rate constant for adsorption is $k_a$ and for desorption $k_d$.

We let $\theta$ be the extent of adsorption which varies from 0 to 1, i.e. how full the surface is, then the rate of change of $\theta$ depends on the rate constant for adsorption , pressure and fraction of vacant sites or $N(1-\theta)$ for N vacant sites. Thus for adsorption $$ \frac{d\theta}{dt} = k_aPN(1-\theta) $$

and for desorption $$ \frac{d\theta}{dt} = -k_dN\theta $$

At equilibrium we make the rate of change equal to zero, i.e. rate of adsorption equals rate of desorption and with a bit of algebra find that $$\theta=\frac{KP}{KP+1}$$ where K is the equilibrium constant or $K=k_a/k_d$ This equation has the fancy name as the Langmuir Isotherm.

You can see that the rate going onto the surface must be equal to the pressure (no gas, no adsorption) and the area free to adsorb onto $(1-\theta)$. (We assume in this model than no molecule double adsorbs; this is dealt with in more advanced treatments). The rate leaving has to be proportional to the amount already there , i.e. proportional to N the total number of surface sites multiplied by the fraction covered.

Ask more questions in the comments if this is not clear.

  • $\begingroup$ I understood Langmuir's isotherm. I didn't understand why rate of adsorption is directly proportional to pressure but rate of desorption is not. $\endgroup$
    – EdmDroid
    Jul 6 '16 at 13:14
  • $\begingroup$ Why is the inverse process concentration-independent? $\endgroup$
    – Marcel
    Jan 3 '17 at 18:27
  • $\begingroup$ @Marcel it has to be proportional to the number on the surface , which is $\theta N$ so is fraction of concentration. No molecules on surface off rate zero, full surface; maximum off rate :) $\endgroup$
    – porphyrin
    Jan 4 '17 at 8:50
  • $\begingroup$ @porphyrin why is desorption not proportional to , say, the decrease of pressure? $\endgroup$ Jan 12 '18 at 18:33
  • $\begingroup$ @sajja islam because we consider adsorption and desorption as separate processes: if there are no molecules on the surface the rate of desorption must be zero and it has to be at a maximum when the surface is completely covered. $\endgroup$
    – porphyrin
    Jan 12 '18 at 22:58

For adsorption, its forward reaction is precursor concentration dependent and the precursor concentration is represented by pressure. For desorption, the forward reaction is also precursor concentration dependent and the concentration here is theta (or one - you can treat the precursor as solid surface so you have concentration of one).


No matter in adsorption or desorption, the believed dominant driving force is the capability for forming or breaking the chemical bond between the adsorbate-surface. Other forces from the environment, such as the van der Waals or Brownian motion are considered negligible during either adsorption or desorption because the activation energy for such processes to occur is low (<20 kJ/mol) but the chemical bonding between adsorbate-surface is usually higher than 200 kJ/mol.

In another word, the chemisorption is the rate-limiting step if one want to calculate the overall rate. Your intuitive feeling that "greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface" is not making sense under Langmuir model because the tendency you mentioned can be only translated into the Brownian motion, which is a less important force.

Under this prerequisite, the overall rate for one given type of molecule/atom to be adsorbed or desorbed on/from the surface can be then calculated and should be highly dependent with the initial concentration of the molecule/atom in either process(e.g. high concentration leads to high chance to be adsorbed). The initial concentration for adsorption is related with the gas pressure while for desorption is related with number of the adsorbates on the surface already.

Therefore, the gas pressure should not affect the desorption process because the rate-limiting step for desorption is only related with how well and fast the chemical bonds between adsorbates-surface would break. The environmental pressure can do nothing about it in theory.

  • $\begingroup$ The scientist's name is van der Waals, not "van der Vaal". Also, logically splitting a substantial amount of text into paragraphs significantly increases readability. $\endgroup$
    – andselisk
    Sep 22 at 11:11

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